[647] leetcode palindromic substring (dynamic programming, String)

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Title Description

1 violence

Complexity Analysis
time complexity: O (n ^ 3)
space complexity: O (1)

class Solution {
public:
    int countSubstrings(string s) {
        if(s.empty()) return 0;
        int ret = 0;
        for(int i = 0; i<s.size(); ++i){
            for(int j = i; j<s.size(); ++j){
                if (isPalindrome(s,i,j))
                    ++ ret;
            }
        }
        return ret;
    }
    // 判断一个字符串是否为回文串
    bool isPalindrome(string s, int i, int j){
        while(i<=j){
            if(s[i++]!=s[j--])
                return false;
        }
        return true;
    }
};

2 from the central character extension

A central palindrome string character position is fixed, may be the center position + palindromic sequence length uniquely represent a palindromic sequence

A bit string to the center, extend to both sides;
respectively count the number of odd and even length of palindromic sequence

Complexity Analysis
time complexity: O (n)
complexity of space: O (1)

class Solution {
public:
    int countSubstrings(string s) {
        if(s.empty()) return 0;
        int ret = 0;
        for(int i = 0; i<s.size(); ++i){
            count(s,i,i,ret);       // 以i位置字符为中心，长度为奇数的回文串数目
            count(s,i,i+1,ret);     // 以i,i+1位置字符为中心，长度为偶数的回文串数目
        }
        return ret;
    }

    void count(string &s, int begin, int end, int &ret){
        while (begin>=0 && end<s.size() && s[begin--] == s[end++])
            ++ ret;
    }
};

3 Dynamic Programming

Two-dimensional array of dp[i][j]string representing s(i,j)whether the string is a palindrome.
(1) First, a string of length 1, 2 need to initialize its value dp
(2) for a length of 3 ~ len

dp[i][j] = (s[i] == s[j] && dp[i+1][j-1])

Note that the above formula when traversing from front to back is not traversed, dp [i] [j] depends on the dp [i + 1] [j -1]. Therefore, the coordinate processing needs to be done, so that the entire process is traversed from front to back.
Because the previously calculated value dp string length 1, so we from i = 3~lentraversing the length of the substring starts, and checks j = 0~len-i+1as a starting point end = j + i -1for the end of the sub-string is palindromic substring.

dp[j][end] = (s[j] == s[end] && dp[j+1][end-1])

Complexity Analysis
time complexity: O (n ^ 2)
Complexity Space: O (n ^ 2)

/*
 * 动态规划
 * 时间复杂度O(n^2) 空间复杂度O(n^2)
 */
class Solution {
public:
    int countSubstrings(string s) {
        if(s.empty()) return 0;
        int len = s.size();
        vector<vector<bool >> dp(len, vector<bool > (len, false));
        int ret = 0;

        // 首先处理长度为1和2的子串
        for (int i = 0; i < len; ++i) {
            dp[i][i] = true;            // 初始化长度为1的回文串
            ++ ret;
        }
        for (int i = 0; i< len -1; ++i){
            if (s[i] == s[i+1]){
                dp[i][i+1] = true;      // 初始化长度为2的回文串
                ++ ret;
            }
        }

        // i为遍历的子串长度,j为起始节点
        for (int i = 3; i <= len ; ++i) {
            for (int j = 0; j <= len -i + 1; ++j) {
                int end = j+i-1;
                if(s[j] == s[end] && dp[j+1][end-1]){
                    dp[j][end] = true;
                    ++ ret;
                }
            }
        }
        return ret;
    }
};